representation of galois group

Galois Representations

These are my live-TeXed notes for the course 18.786: Galois Representations taught by Sug Woo Shin at MIT, Spring 2014. References .

Any mistakes are the fault of the notetaker. Let me know if you notice any mistakes or have any comments!

Deformations of Galois representations

Group-theoretic hypothesis.

• Use Burnside basis theorem (see [Boe] Ex 1.8.1).

Liftings of mod representations

Some background on irreducible representations, deformations of mod representations, linear algebraic lemmas, tangent spaces, generic fiber of universal lifting rings.

The following is a basic algebraic fact.

Deformation problems

So one can work directly at the level of cohomology instead of cocycles.

Global Galois deformation problems

We begin with a remark on fixing the determinant.

Next we will define a deformation functor and show it is representable and study its ring theoretic properties.

Presenting global deformation rings over local lifting rings

Two modifications are needed in general:

This motivates the definition of the mysterious complex in [Gee].

• the usual local and global Galois cohomology,
• the dimension of the "dual Selmer group" (as the error term).

Computation of

Assume for simplicity that

To execute the last two steps, We need the following facts.

From the long exact sequence in cohomology, it follows that

Another input is the determination of the Euler-Poincare characteristics.

The final key inputs are the local and global duality theorems.

Notice that 1, 4, 5 terms are the same as in Poitou-Tate.

(I was out of town for AWS 2014, this section is shameless copied from Rong Zhou's typed notes.)

Local universal lifting rings

• Each component is formally smooth.

The Weil-Deligne functor

Hodge-tate weights, potentially semistable local lifting rings.

Galois Representations

These are a blog-post form of notes I wrote to introduce the machinery and motivate the study of Galois representations.

Introduction

Considering the importance of Galois groups in number theory and geometry, it is natural to study their representation theory. However, Galois groups have additional structure which makes the theory of their representations remarkably rich. They are profinite topological groups and using topological arguments is extremely fruitful in studying their general representation theory. We will see that the topology is too restrictive to admit interesting Galois representations over $\mathbb{C}$. However, the profinite topology interacts much more favorably with $\ell$-adic numbers due to their profinite-like topology. Therefore, we will seek out vector spaces over $\mathbb{Q}_{\ell}$ on which the Galois group naturally acts. Furthermore, Galois groups act on algebraic fields and preserve certain polynomial equations meaning that Galois groups act naturally on algebraic varieties built from fields and polynomials. This action allows many Galois representations to be viewed as automorphisms of certain geometric objects giving a powerful link between the number theory of field extensions and the geometry of algebraic objects.

Let $G$ be a profinite group and $F$ a topological field. An $n$-dimensional representation of $G$ is a continuous homomorphism, \(\rho : G \to \mathrm{GL}_{n}(F)\) If $G = G_K = \mathrm{Gal} \small(\bar{K} / K \small)$, the absolute Galois group of $K$, then we call such a representation a Galois representation and if $F$ is algebraic over $\mathbb{Q}_{\ell}$ then we call it an $\ell$-adic Galois representation.

Projective Limits

A projective system is a family of objects indexed by a poset $(I, \le)$ with morphisms $f_{ij} : A_j \to A_i$ when $i \le j$ such that,

$f_{ii} = \mathrm{id}_{A_i}$

$f_{ik} = f_{ij} \circ f_{jk}$ for all $i \le j \le k$

$(I, \le)$ is directed meaning that for every $i, j \in I$ there exists $k \in I$ such that $i \le k$ and $j \le k$. This means that for all $A_i$ and $A_j$ there is an object $A_k$ such that there are maps $f_{ik} : A_k \to A_i$ and $f_{jk} : A_k \to A_j$.

we define the projective limit $\varprojlim A_n$ to be the categorical limit of this system. Concretely, for groups or modules, we can give the explicit construction of such an object,

Therefore, the projective limit is the set of sequences which reduce compatibly under the maps $f$.

A very important special case is that of a leftward mapping sequence where $I = \mathbb{N}$ with the usual order.

Given a diagram,

we define the projective limit $\varprojlim A_n$ to be the categorical limit of the diagram. Concretely, for groups or modules, we can give the explicit construction of such an object, \(\varprojlim A_n = \left\{ (a_n) \in \prod_n A_n \: \bigg| \: \forall n \in \mathbb{N}: f_n(a_n) = a_{n-1} \right\}\) Therefore, the projective limit is the set of sequences which reduce compatibly under the maps $f$. :::

::: remark Remark 1 . One should view the projective limit as the object which “naturally projects” compatibly with the maps onto each of the given objects. There are clear projection maps $\pi_i : \varprojlim A_n \to A_i$ given by $\pi_i((a_n)) = a_i$. Reversing all the maps, we can define the dual notion called the direct limit which is the objective into which each of the given objects include compatibly via maps $\iota_i : A_i \to \varinjlim A_n$. When the given morphisms are inclusions the direct limit is simply the union. :::

::: example Example 1 . Let $R$ be a ring. The ring of formal power series on $R$ is \(R[[X]] \cong \varprojlim R[X] / X^n R[X]\) with maps $R[X] / X^{n+1} R[X] \to R[X] / X^n R[X]$ given by reduction modulo $X^n$. The sequences making up the projective limit give the partial sums of a formal power series. :::

Infinite Galois Theory

::: proposition Proposition 1 . Let $F / K$ be Galois. Then there exists an isomorphism, \(\mathrm{Gal} \small(F / K \small) \cong \varprojlim_{L/K} \mathrm{Gal} \small(L / K \small)\) where $L$ runs over all finite Galois extensions $K \subset L \subset F$. The projective system is given by the restriction maps $\mathrm{Gal} \small(L / K \small) \to \mathrm{Gal} \small(L’ / K \small)$ when $L’ \subset L$. :::

::: proof Proof. Given $\sigma \in \mathrm{Gal} \small(F / K \small)$ we consider the restriction $\sigma |_L$ to each finite Galois extension $L / K$ which are clearly compatible with restrictions between finite extensions. This gives a map to the projective limit. Since each $\alpha \in F$ is algebraic over $K$ we know that $\alpha$ lies in a finite Galois extension of $K$ so if $\sigma$ is trivial on all finite Galois extensions then $\sigma(\alpha) = \alpha$ so $\sigma = \mathrm{id}_F$. Thus the map is injective. Furthermore, an element of the projective limit induces an automorphism of $F / K$ by mapping each $\alpha \in F$ to its image under the automorphism acting on any finite Galois extension containing $\alpha$. Thus the mapping is surjective. ◻ :::

::: remark Remark 2 . The above identification gives a natural profinite topology on $\mathrm{Gal} \small(F/ K \small)$ by making the projection maps $\mathrm{Gal} \small(F / K \small) \to \mathrm{Gal} \small(L / K \small)$ continuous for each finite Galois extension $L / K$. In particular, the kernels of these maps $\mathrm{Gal} \small(F / L \small)$ are open subgroups and form a neighborhood basis of $\mathrm{id}$. :::

::: example Example 2 . The absolute Galois group of $\mathbb{F}_{p}$ is equal to the profinite completion of $\mathbb{Z}$, \(\mathrm{Gal} \small(\overline{\mathbb{F}_{p}} / \mathbb{F}_{p} \small) \cong \hat{\mathbb{Z}} = \varprojlim \mathbb{Z}/ n \mathbb{Z}\) :::

::: theorem Theorem 2 (Galois Correspondence). Let $F / K$ be a Galois extension with $G = \mathrm{Gal} \small(F / K \small)$. There is an inclusion reversing correspondence between closed subgroups $H \subset G$ and subfields $K \subset L \subset F$ given by $H \mapsto F^H$ and $L \mapsto \mathrm{Gal} \small(F / L \small)$. Furthermore, finite extensions $K \subset L \subset F$ correspond to open subgroups $\mathrm{Gal} \small(F / L \small) \subset G$ whose cosets correspond to embeddings of $L$ into $F$ fixing $K$. Galois extensions $K \subset L \subset F$ correspond to closed normal subgroups. :::

$\ell$-adic Numbers

::: trivlist The $\ell$-adic integers are the projective limit, \(\mathbb{Z}_{\ell} = \varprojlim \mathbb{Z}/ \ell^n \mathbb{Z}\) under the maps $\mathbb{Z}/ \ell^{n+1} \mathbb{Z}\to \mathbb{Z}/ \ell^n \mathbb{Z}$ given by reduction mod $\ell$. :::

::: remark Remark 3 . There is an inclusion $\mathbb{Z}\hookrightarrow \mathbb{Z} {\ell}$ given by reducing $a \in \mathbb{Z}$ modulo each $\ell^n$. The sequences representing $\mathbb{Z}\subset \mathbb{Z} {\ell}$ are exactly those which are eventually constant after the largest power dividing the integer in question. Using the intuition gained from the ring of formal power series, we can write any $\ell$-adic integer as a formal “base-$\ell$” power series, \(z = a_0 + a_1 \ell + a_2 \ell^2 + a_3 \ell^3 + \cdots\) which is the natural extension of how integers may be represented in base $\ell$. Although this expression is simply convenient and suggestive shorthand for the projective limit sequence of partial sums, \(z = (a_0, \, a_0 + a_1 \ell, \, a_0 + a_1 \ell + a_2 \ell^2, \, a_0 + a_1 \ell + a_2 \ell^2 + a_3 \ell^3, \, \cdots)\) we can actually give meaning to this infinite sum by changing the standard definition of convergence. Define the $\ell$-adic valuation $v_{\ell} : \mathbb{Z} {\ell} \to \mathbb{N}\cup { \infty }$ by $v {\ell}((a_i))$ equals the index of the first nonzero term $a_i$ and $v_{\ell}(0) = \infty$. All terms in the sequence past $v_{\ell}((a_i))$ are nonzero because if $a_i = 0$ then $a_{i-1} = f_i(a_i) = 0$. We can then define an absolute value, $|z| {\ell} = \ell^{-v {\ell}(z)}$ which gives a non-archimedean metric on $Z_{\ell}$. Under this metric, the sequence of elements in $\mathbb{Z}\subset \mathbb{Z} {\ell}$ given by these the partial sums actually does converge to the $\ell$-adic number, \(z = a_0 + a_1 \ell + a_2 \ell^2 + a_3 \ell^3 + \cdots\) because the element, \(z - z_{N-1} = z - \sum_{i = 0}^{N-1} a_i \ell^n = a_N \ell^N + a_{N+1} \ell^{N+1} + a_{N+2} \ell^{N+2} + \cdots\) has valuation $v \ell(z_n) \ge N$ because $z_N = (0, \cdots, a_N \ell^N, a_{N+1} \ell^{N+1}, \cdots)$ where the first $N-1$ terms are $0$. Therefore, \(|z - z_n|_{\ell} \le \frac{1}{\ell^N} \to 0\) so the sequence converges $z_n \to z$. :::

::: trivlist The $\ell$-adic field is the field of fractions of $\mathbb{Z} {\ell}$, \(\mathbb{Q}_{\ell} = \mathrm{Frac}(\mathbb{Z}_{\ell})\) on which we extend the $\ell$-adic valuation to $v {\ell} : \mathbb{Q} {\ell} \to \mathbb{Z}$ by $v {\ell}(a/b) = v_{\ell}(a) - v_{\ell}(b)$. :::

::: proposition Proposition 3 . $\mathbb{Z} {\ell}^\times = { z \in \mathbb{Z} {\ell} \mid |z|_{\ell} = 1 }$ :::

::: proof Proof. Let $z = (a_i) \in \mathbb{Z} \ell$. If $v {\ell}(z) > 0$ then $a_0 = 0$ so for any $(b_i) \in \mathbb{Z} {\ell}$ we have $a_0 b_0 = 0$ so $z \notin \mathbb{Z} {\ell}$. However, if $v_{\ell}(z) = 0$ then choose $b_n = a_n^{-1} \in \mathbb{Z}/ \ell^n \mathbb{Z}$ which exists because $a_n$ is coprime to $\ell$ since it projects down to $a_0 \neq 0$ in $\mathbb{Z}/ \ell \mathbb{Z}$. Then we have, \((a_i) \cdot (b_i) = (a_i b_i) = (1)\) so $z \in \mathbb{Z}_{\ell}^\times$. ◻ :::

::: proposition Proposition 4 . Every element $z \in \mathbb{Q} {\ell}$ can be written uniquely as $z = \ell^n u$ where $u \in \mathbb{Z} {\ell}^\times$ and $n = v_{\ell}(z)$. :::

::: proof Proof. First we will prove this for $z = (a_i) \in \mathbb{Z} {\ell}$. Take $n = v {\ell}(z)$ so we know that $f_{n-1, k}(a_{k}) = 0$ so $\ell^n \mid a_k$ but $\ell^{n+1}$ does not. Thus we can write $a_k = \ell^n u_k$ with $u_k \in (\mathbb{Z}/ \ell^k \mathbb{Z})^\times$. Take $u = (u_k)$ with $u_k = f_{k n}(u_n) \neq 0$ since $\ell \centernot \mid u_n$ for $k < n$ so clearly $z = \ell^n u$ and $v_{\ell}(u) = 0$. Furthermore, if $z \in \mathbb{Q} {\ell}$ then $z = \frac{a}{b}$ for $a,b \in \mathbb{Z} {\ell}$ so we can write $a = \ell^n u$ and $b = \ell^m v$ with $u,v \in \mathbb{Z}^\times$. Thus, \(z = \frac{\ell^n u}{\ell^m v} = \ell^{n-m} u v^{-1}\) with $v_{\ell}(z) = v_{\ell}(a) - v_{\ell}(b) = n - m$ an $u v^{-1} \in \mathbb{Z}_{\ell}^\times$. ◻ :::

::: proposition Proposition 5 . \(\mathbb{Q}_{\ell} = \mathbb{Z}_{\ell}\left[ \frac{1}{\ell} \right] = \bigcup_n \frac{1}{\ell^n} \mathbb{Z}_{\ell}\) Therefore we may represent an element of $\mathbb{Q}_{\ell}$ as a power series, \(a_{-N} \ell^{-N} + a_{-N + 1} \ell^{-N + 1} + \cdots + a_0 + a_1 \ell + a_2 \ell^2 + \cdots\) with only finitely many negative exponent terms. :::

::: proof Proof. By the previous proposition we can write any element $z \in \mathbb{Q} {\ell}$ as $\ell^{n} u$ for $u \in \mathbb{Z} {\ell}^\times$ and $n \in \mathbb{Z}$. Therefore, we simply need to invert $\ell$ to get negative powers of $\ell$ to represent all of $\mathbb{Q} {\ell}$ from $\mathbb{Z} {\ell}$. ◻ :::

::: proposition Proposition 6 . $\mathbb{Z} {\ell}$ is a local PID (and thus a discrete valuation ring) with unique maximal ideal $\ell \mathbb{Z} {\ell}$ and residue field $\mathbb{F}_{\ell}$. :::

::: proof Proof. Let $I \subset \mathbb{Z} {\ell}$ be an ideal. Consider $n = v {\ell}(I) = \min { v_{\ell}(z) \in \mathbb{N}\mid z \in I }$ where the minimum value exists by well ordering. Thus, there exists $z_0 \in I$ with $v_{\ell}(z_0) = n$. I claim that $I = (\ell)^n$. We can write $z_0 = \ell^n u$ for some $u \in \mathbb{Z} {\ell}^\times$. Therefore, $I \subset (z_0) = (\ell^n u) = (\ell)^n$. Furthermore for any $z \in I$ we have $v {\ell}(z) = m \ge n$ so $\ell^n \mid z$ since $z = \ell^m v$ for $v \in \mathbb{Z} {\ell}^\times$ so $z = \ell^{m - n} v \ell^n$ with $\ell^{m - n} v \in \mathbb{Z} {\ell}$ and thus $z \in (\ell)^n$. Therefore, $I = (\ell)^n$. Thus, all proper ideals are contained in $(\ell)$ so $\mathfrak{m} = (\ell)$ is the unique maximal ideal. Consider the map, \(\phi : \mathbb{Z}\to \mathbb{Z}_{\ell} / \ell \mathbb{Z}\) given by inclusion and then projection. Given $z = (a_i) \in \mathbb{Z} {\ell}$ take $a \in \mathbb{Z}$ such that $a \equiv a_i \; \mathrm{mod} \; {\ell}$. Then $v {\ell}(z - a) \ge 1$ so $\ell \mid z - a$. Therefore, $[a] = [z]$ in $\mathbb{Z} {\ell} / \ell \mathbb{Z} {\ell}$ so $\phi$ is surjective. Furthermore, $\ker{\phi} = \ell \mathbb{Z}$ since $[a] = 0$ exactly when $a = (0, a, \cdots)$ i.e. $a \equiv 0 \; \mathrm{mod} \; \ell$. Therefore, \(\mathbb{Z}/ \ell \mathbb{Z}\cong \mathbb{Z}_{\ell} / \ell \mathbb{Z}\) so the residue field is given by $\mathbb{Z} {\ell} / \mathfrak{m} \cong \mathbb{F} {\ell}$. ◻ :::

::: proposition Proposition 7 . $\mathbb{Q} {\ell} / \mathbb{Z} {\ell} \cong \mathbb{Q}/ \mathbb{Z}$ and $\mathbb{Z} {\ell} / \ell^n \mathbb{Z} {\ell} \cong \mathbb{Z}/ \ell^n \mathbb{Z}$ :::

::: proof Proof. Quotienting by $\ell^n \mathbb{Z}_{\ell}$ is equivalent to ignoring all elements of the sequence with index greater than or equal to $n$. Therefore, we can choose a rational number (or integer) which reduces modulo $\ell^{n} \mathbb{Z}$ to the required value which is consistently reduced by the reduction maps. ◻ :::

::: trivlist A complete non-archimedean field $K$ is a topological field which is complete with respect to an absolute value satisfying the non-archimedean property or ultrametric inequality, \(|\alpha + \beta| \le \max\{|\alpha|, |\beta|\}\) For example $\mathbb{Q} {\ell}$ with the $\ell$-adic absolute value $| \cdot | : \mathbb{Q} {\ell} \to \mathbb{Z}$. :::

::: proposition Proposition 8 . Let $K$ be a complete non-archimedean field and $L / K$ a separable extension. The absolute value on $K$ extends uniquely to a non-archimedean absolute value on $L$. Furthermore, if $L / K$ is finite then $L$ is complete with respect to the extended absolute value. :::

::: lemma Lemma 9 (Krasner). Let $K$ be a complete non-archimedean field and $\alpha, \beta \in \bar{K}$. If $\alpha$ is strictly closer to $\beta$ than to any conjugate of $\alpha$ then $K(\alpha) \subset K(\beta)$. :::

::: proof Proof. Consider an automorphism $\sigma \in \mathrm{Gal} \small(\bar{K} / K \small)$. By assumption, $|\alpha - \beta | < |\alpha - \sigma(\alpha)|$ whenever $\sigma(\alpha) \neq \alpha$. Suppose that $\sigma(\beta) = \beta$ and consider the value, \(|\alpha - \sigma(\alpha)| = |\alpha - \beta + \beta - \sigma(\alpha) | \le \max \left\{ | \alpha - \beta |, | \beta - \sigma(\alpha)| \right\}\) We know that $|\beta - \sigma(\alpha)| = |\sigma(\beta - \alpha)| = |\alpha - \beta|$ by uniqueness of the absolute value. Therefore, unless $\sigma(\alpha) = \alpha$, \(|\alpha - \sigma(\alpha)| \le |\alpha - \beta| < |\alpha - \sigma(\alpha)|\) which is a contradiction so $\sigma(\alpha) = \alpha$. Therefore, $\mathrm{Gal} \small(\bar{K} / K(\beta) \small) \subset \mathrm{Gal} \small(\bar{K} / K(\alpha) \small)$ and thus $K(\alpha) \subset K(\beta)$. ◻ :::

::: {#thm:algebraic_extensions_adics .theorem} Theorem 10 . Let $K / \mathbb{Q} {\ell}$ be finite. There exists $\alpha \in \overline{\mathbb{Q}}$ such that $K = \mathbb{Q} {\ell}(\alpha)$. :::

::: proof Proof. By the primitive element theorem, $K = \mathbb{Q} {\ell}(\alpha’)$ for some $\alpha’$ algebraic over $\mathbb{Q} {\ell}$ with minimal polynomial $f \in \mathbb{Q} {\ell}[X]$. Take $g \in \mathbb{Q}[X]$ which is monic of the same degree. Write, \(g(X) = \prod_{i = 0}^n (X - \alpha_i)\) for roots $\alpha_i \in \overline{\mathbb{Q}}$. Consider, \(|(f - g)(\alpha')|_{\ell} = |g(\alpha')|_{\ell} = \prod_{i = 0}^n |\alpha' - \alpha_i|_{\ell}\) by choosing $g$ such that $f - g$ is sufficiently small we can ensure that for any $\epsilon > 0$ there is some root $\alpha$ of $g$ such that $|\alpha’ - \alpha| < \epsilon$. In particular, for sufficiently small $\epsilon$ the root $\alpha$ will be strictly closer to $\alpha’$ than any conjugate of $\alpha’$. Therefore, by Krasner’s Lemma, \(\mathbb{Q}_{\ell}(\alpha') \subset \mathbb{Q}_{\ell}(\alpha)\) However because $f$ is irreducible, \([\mathbb{Q}_{\ell}(\alpha) : \mathbb{Q}_{\ell}] \le \deg{g} = \deg{f} = [\mathbb{Q}_{\ell}(\alpha') : \mathbb{Q}_{\ell}] \le [\mathbb{Q}_{\ell}(\alpha) : \mathbb{Q}_{\ell}]\) forcing an equality. Therefore $\mathbb{Q} {\ell}(\alpha) = \mathbb{Q} {\ell}(\alpha’) = K$ and furthermore $g$ is irreducible in $\mathbb{Q} {\ell}[X]$ thus in $\mathbb{Q}[X]$ so, \([\mathbb{Q}(\alpha) : \mathbb{Q}] = \deg{g} = [\mathbb{Q}_{\ell}(\alpha) : \mathbb{Q}_{\ell} ]\) ◻ :::

Galois Representations over $\mathbb{C}$

Lemma 11 . There exists a neighborhood $V$ of $I$ in $\mathrm{GL}_{n}(\mathbb{C})$ that contains no nontrivial subgroup. :::

::: proof Proof. Recall that $M_n(\mathbb{C})$ is a metric space under the absolute value $|A| = \max |A_{ij}|$. Let $U_r = {A \in M_n(\mathbb{C}) : |A - I| < r \text{ and } \mathrm{Tr} ! \left(A\right) = 0 }$ and take $V_r = \exp(U_r)$ an open neighborhood of $I \in \mathrm{GL}_{n}(\mathbb{C})$ since $\det{\exp{A}} = \exp{\mathrm{Tr} ! \left(A\right)} = 1$. Suppose that $H \subset V_r$ is a subgroup. For $B \in H$ we have $B = \exp{A}$ and thus $B^k = (\exp{A})^k = \exp{(kA)}$ so $kA \in U_r$. However, $|k A| = |k| \cdot |A|$ which, by the archimedean property, can be taken arbitrarily large if $|A| > 0$. Since all $A \in U_r$ have $|A| < r$ this contradicts the fact that $k A \in U_r$ unless $|A| = 0 \implies A = 0 \implies B = I$. Thus, $H = { I }$. ◻ :::

::: remark Remark 4 . The above proof depends crucially on the archimedean property. :::

::: proposition Proposition 12 . Any continuous homomorphism $\rho: G_{K} \to \mathrm{GL}_{n}(\mathbb{C})$ factors through $\mathrm{Gal} \small(F/K \small)$ for some finite Galois extension $F / K$. Hence its image is finite. :::

::: proof Proof. By Lemma 11 {reference-type=”ref” reference=”lem:no_subgroups_of_nbd”}, let $V$ be an neighborhood of $I$ in $\mathrm{GL}_{n}(\mathbb{C})$ which contains no non-trivial subgroups. Then $U = \rho^{-1}(V)$ is an open neighborhood of $\mathrm{id}\in G_K$ and thus contains a normal subgroup of the form $\mathrm{Gal} \small(\bar{K} / F \small)$ for some galois extension $F / K$. Since $\rho$ is a homomorphism, the image of $\mathrm{Gal} \small(\bar{K} / F \small)$ is subgroup contained in $V$. But $V$ does not have any nontrivial subgroup so $\mathrm{Gal} \small(\bar{K} / F \small) \subset \ker{\rho}$ is actually in the kernel of $\rho$. Thus, $\rho$ factors through the quotient, \(\mathrm{Gal} \small(\bar{K} / K \small) / \mathrm{Gal} \small(\bar{K} / F \small) \cong \mathrm{Gal} \small(F / K \small)\) which is finite. Hence $\rho$ has finite image. ◻ :::

::: corollary Corollary 13 . Every $1$-dimensional Galois representation of $G_{\mathbb{Q}}$ over $\mathbb{C}$ is a Dirichlet character. :::

::: proof Proof. Any Galois representation $\rho : G_{\mathbb{Q}} \to \mathrm{GL}_{1}(\mathbb{C}) \cong \mathbb{C}^\times$ factors through the quotient, \(\rho : \mathrm{Gal} \small(F / \mathbb{Q} \small) \to \mathbb{C}^\times\) where $F$ is a Galois number field. However, $\mathbb{C}^\times$ is abelian so $\rho$ descends to the abelianization. Let $C \triangleleft \mathrm{Gal} \small(F / \mathbb{Q} \small)$ be the commutator subgroup. By the Galois correspondence, $F^{\mathrm{ab}} = F^C$ is a Galois extension of $\mathbb{Q}$ with Galois group, \(\mathrm{Gal} \small(F^{\mathrm{ab}} / \mathbb{Q} \small) \cong \mathrm{Gal} \small(F / \mathbb{Q} \small) / C = \mathrm{Gal} \small(F / \mathbb{Q} \small)^{\mathrm{ab}}\) By the Kronecker-Weber theorem, every abelian extension of $\mathbb{Q}$ lies inside some cyclotomic field. So $F^{\mathrm{ab}} \subset \mathbb{Q}(\zeta_N)$ giving a restriction map on the Galois groups. This gives a commutative diagram,

::: center :::

Thus $\rho$ factors through a Dirichlet character, \(\chi : \mathrm{Gal} \small( \mathbb{Q}(\zeta_n) / \mathbb{Q} \small) \cong (\mathbb{Z}/ N \mathbb{Z})^\times \to \mathbb{C}^\times\) ◻ :::

$\ell$-adic Galois Representations

The archimedean nature of $\mathbb{C}$ leading to Lemma 11 {reference-type=”ref” reference=”lem:no_subgroups_of_nbd”} made the theory of complex Galois representations fairly uninteresting. However, if we consider a non-archimedean field such as $\mathbb{Q}_{\ell}$, this restriction is lifted. In fact,

::: proposition Proposition 14 . Every neighborhood of $1 \in \mathbb{Q}_{\ell}^\times$ contains a nontrivial subgroup. :::

::: proof Proof. Let $U$ be an open neighborhood of $1 \in \mathbb{Q} {\ell}^\times$, then there exists $n \in \mathbb{Z}^+$ such that \(V(n) = 1 + \ell^n \mathbb{Z}_{\ell} \subset U\) However, $V(n)$ is a nontrivial subgroup of $\mathbb{Q} {\ell}^\times$ because \((1 + \ell^n z)^{-1} - 1 = \frac{ \ell^n z }{1 + \ell^n z} = \ell^n \frac{z}{1 + \ell^n z} \in \ell^n \mathbb{Z}_{\ell}\) since $1 + \ell^n z \in \mathbb{Z}_{\ell}^\times$. ◻ :::

One-Dimensional Galois Representations

::: trivlist Let $F/\mathbb{Q}$ be Galois, $p \in \mathbb{Z}$ be a prime, and $\mathfrak{p}$ is a prime in $\mathcal{O} {F}$ lying above $p$. Then $\mathrm{Frob} {\mathfrak{p}}\in \mathrm{Gal} \small(F/\mathbb{Q} \small)$ is defined by \(\mathrm{Frob}_{\mathfrak{p}}(x) \equiv x^p \: \: \mathrm{mod} \: \mathfrak{p}\) for $x \in \mathcal{O}_{F}$. :::

::: lemma Lemma 15 . Let $F = \mathbb{Q}(\zeta_N)$. Let $p$ be a prime in $\mathbb{Z}$ such that $p \centernot \mid N$. Let $\mathfrak{p}$ be a prime in $F$ lying above $p$. Then $\mathfrak{f} {\mathfrak{p}} = \mathcal{O} {F} / \mathfrak{p} = \mathbb{F} {p}[\zeta_N]$. Let $x \in \mathcal{O} {F}$, then we can describe the action of $\mathrm{Frob} {\mathfrak{p}}$ by \(\mathrm{Frob}_{\mathfrak{p}}\left( \sum_{i = 0}^{N-1} a_i \zeta_N^i \right) \equiv \left( \sum_{i = 0}^{N-1} a_i \zeta_N^i \right)^p \equiv \sum_{i = 0}^{N-1} a_i \zeta_N^{ip} \: \: \mathrm{mod} \: \mathfrak{p}\) That is to say, the action of $\mathrm{Frob} {\mathfrak{p}}$ takes $\zeta_N$ to $\zeta_N^p$. :::

::: trivlist The $\ell$-adic cyclotomic character $\chi_{\ell} : G_\mathbb{Q}\to \mathbb{Q} {\ell}^\times$ of $G \mathbb{Q}$ is defined by, \(\sigma \mapsto (m_1, m_2, m_3, \dots ) \text{ where } \sigma(\zeta_{\ell^n}) = \zeta_{\ell^n}^{m_n}\) is a $1$-dimensional Galois representation since $m_n$ is defined up to multiples of $\ell^n$. :::

::: remark Remark 5 . Notice that when $p \neq \ell$ we have, \(\chi_{\ell}(\mathrm{Frob}_{\mathfrak{p}}) = p\) In particular, the image of $\chi_{\ell}$ is infinite. Therefore, $\ell$-adic Galois representations allow for richer structure than those over $\mathbb{C}$. :::

::: trivlist We write $\mathbb{Q} \ell(1)$ for the vectorspace $\mathbb{Q} \ell$ with the cyclotomic character Galois rep $\chi_\ell : G_\mathbb{Q}\to \mathbb{Q} \ell^\times = \mathrm{GL} {1}(\mathbb{Q} \ell)$ acting on it. For any $\mathbb{Q} \ell$-rep of $G$ we define the $m^{\text{th}}$-Tate twist via, \(V(m) = V \otimes_{\mathbb{Q}_\ell} \mathbb{Q}_\ell(1)^{\otimes m} = V \otimes_{\mathbb{Q}_\ell} \mathbb{Q}_\ell(m)\) and we define, \(\mathbb{Q}_\ell(-1) = \mathbb{Q}_\ell(1)^{\otimes -1} = \mathbb{Q}_{\ell}(1)^\vee\) the dual representation. :::

::: remark Remark 6 . Note that there is a more natural way to see the cyclotomic character arrise. The torsions points $\mu_{\ell^n}(\overline{\mathbb{Q}}) = \overline{\mathbb{Q}}[\ell^n]$ have a natural action of $G_{\mathbb{Q}}$ which is compatible with restriction maps. Then we define, \(\mathbb{Z}_\ell(1) = \varprojlim_{n} \mu_{\ell^n}(\overline{\mathbb{Q}})\) How does $G_\mathbb{Q}$ act on $\mathbb{Z} \ell(1)$. Well, we take a sequence $(\zeta \ell^{a_1}, \zeta_{\ell^2}^{a_2}, \zeta_{\ell^3}^{a_3}, \dots )$ and map it to, \(\sigma(\zeta_\ell^{a_1}, \zeta_{\ell^2}^{a_2}, \zeta_{\ell^3}^{a_3}, \dots ) = (\sigma(\zeta_\ell^{a_1}), \sigma(\zeta_{\ell^2}^{a_2}), \sigma(\zeta_{\ell^3}^{a_3}), \dots ) = (\zeta_\ell^{m_1 a_1}), \zeta_{\ell^2}^{m_2 a_2}, \zeta_{\ell^3}^{m_3 a_3}, \dots )\) so representing the sequence by its exponents, $\sigma \in G_\mathbb{Q}$ acts via multiplication by the sequence $(m_1, m_2, m_3, \dots)$. Then we can define the Galois-rep $\mathbb{Q} \ell(1)$ as, \(\mathbb{Q}_\ell(1) = \mathbb{Z}_\ell(1) \otimes_{\mathbb{Z}_\ell} \mathbb{Q}_\ell\) where $\mathbb{Q} \ell$ is the trivial $\ell$-adic $G$-rep and we take the tensor product as $G$-modules. We can do the same construction for all primes simultaneously to get, \(\hat{\mathbb{Z}}(1) = \varprojlim_n \mu_n(\overline{\mathbb{Q}})\) which has an analogous $G_\mathbb{Q}$-action. In fact, by the Chinese Remainder Theorem, there is a canonical decomposition of $G_\mathbb{Q}$-modules, \(\hat{\mathbb{Z}}(1) = \prod_{\ell} \mathbb{Z}_\ell(1)\) However, by the above product decomposition, $\hat{\mathbb{Z}}(1)$ is not a domain so we cannot make it into a field and discuss its corresponding representations which is why we content ourselves to work $\ell$-adically in each prime. :::

Higher-Dimensional Galois Representations

::: theorem Theorem 16 . Let $G$ be profinite. Any continuous homomorphism $\rho : G \to \mathrm{GL} {n}(\overline{\mathbb{Q} {\ell}})$ has image $\rho(G) \subset \mathrm{GL} {n}(K)$ where $K$ is a finite extension of $\mathbb{Q} {\ell}$. :::

::: proof Proof. By Theorem 10 {reference-type=”ref” reference=”thm:algebraic_extensions_adics”}, the finite extensions of $\mathbb{Q} {\ell}$ are indexed by elements $\alpha \in \overline{\mathbb{Q}}$ and are thus countable. Write, \(\overline{\mathbb{Q}_{\ell}} = \bigcup_{i \in I} E_i\) where $E_i$ are finite extensions of $\mathbb{Q} {\ell}$ and $I$ is countable. A topological space $X$ is a Baire space if and only if given any countable collection of closed sets $F_i$ in $X$, each with empty interior in $X$, their union also has empty interior. The image $\rho(G)$ is compact because $G$ is compact and $\rho$ is continuous. Therefore $\rho(G)$ is a complete metric space and thus Baire space by the Baire category theorem. Let $F_i$ be the closure of $\mathrm{GL} {n}(E_i) \cap \rho(G)$ in $\rho(G)$. Then \(\rho(G) = \bigcup_{i \in I} F_i\) has nonempty interior in $\rho(G)$. Therefore, there exists $i \in I$ such that $F_i$ contains a non-empty open subset $U \subset \rho(G)$. After translating and shrinking $U$, we may assume it is an open subgroup of $\rho(G)$. The quotient $\rho(G) / U$ is covered by the sets $\mathrm{GL} {n}(E_j) \cap \rho(G)$ for each $E_j \supset E_i$. Since $\rho(G)$ is compact, the cover by open cosets must be finite because there do not exist any proper subcovers. Thus the quotient $\rho(G) / U$ is finite so we need only a finite number of such $j$. The compositum $K$ of the fields $E_j$ is then a finite extension of $\mathbb{Q} {\ell}$. Furthermore, $\rho(G) \subset \mathrm{GL} {n}(K)$ because it covers $\rho(G) / U$ and $U \subset \mathrm{GL} {n}(K)$ so each coset $g U \subset \mathrm{GL} {n}(K)$. ◻ :::

::: corollary Corollary 17 . All $\ell$-adic Galois representations $\rho : \mathrm{Gal} \small(F / K \small) \to \mathrm{GL} {n}(\overline{\mathbb{Q} {\ell}})$ are of the form, \(\rho : \mathrm{Gal} \small(F / K \small) \to \mathrm{GL}_{n}(\mathbb{Q}_{\ell}(\alpha))\) where $\alpha \in \overline{\mathbb{Q}}$. :::

Galois Representations Attached to Elliptic Curves

::: trivlist An elliptic curve is a smooth, projective curve of genus one with a distinguished point $O$. More concretely, an elliptic curve $E$ defined over a field $K$ is a planar algebraic curve given by, \(\left\{ (x,y) \in \bar{K}^2 \mid y^2 = x^3 + ax^2 + bx + c \right\} \cup \left\{ O \right\} \subset \bar{K}^2\) for constants $a,b,c \in K$ such that the equation is non-singular. The distinguished point $O$ is viewed as the “point at infinity.” For any $L / K$, the set of $L$-rational points of $E$ is, \(E(L) = \left\{ (x,y) \in L^2 \mid y^2 = x^3 + ax^2 + bx + c \right\} \cup \left\{ O \right\} \subset L^2\) When $L/K$ is algebraic, that is, $K \subset L \subset \bar{K}$, then $E(L) = E \cap L$ which is exactly the set of fixed points of $E$ under the action of $\mathrm{Gal} \small(\bar{K}/L \small)$. :::

::: remark Remark 7 . Elliptic curves are, remarkably, abelian varieties meaning there exists a map $+ : E \times E \to E$ expressed by rational functions which gives $E$ the structure of an abelian group with identity $O$. :::

::: trivlist Let $E$ be an elliptic curve defined over $K$. For $n \in \mathbb{Z}$, define the $n$-torsion of $E$, denoted by $E[n]$, to be the kernel of the map $x \mapsto n x$. :::

::: proposition Proposition 18 . Let $E$ be an elliptic curve defined over a number field $K$ (in particular $\mathbb{Q}$) then $E[n] \cong (\mathbb{Z}/ n \mathbb{Z}) \oplus (\mathbb{Z}/ n \mathbb{Z})$ :::

::: proof Proof. I will give a sketch of the proof assuming some knowledge of Weierstrass elliptic functions. Associated to any elliptic curve $E$ over $\mathbb{C}$ is a complex lattice $\Lambda \subset \mathbb{C}$ and a doubly periodic meromorphic function $\wp(z ; \Lambda)$ defined on $\mathbb{C}$ which is constant on translates by $\Lambda$. Therefore $\wp$ descends to a function of quotient $\mathbb{C}/ \Lambda \to \mathbb{C}$. It turns out that, magically, $\wp : \mathbb{C}/ \Lambda \to E$ given by $z \mapsto (\wp(z), \wp’(z))$ is an isomorphism of groups. In particular, $E$ is topologically a torus. Let $\Lambda$ be generated by two complex numbers $\omega_1, \omega_2$ called the fundamental periods of the lattice. Then the $n$-torsion points in $\mathbb{C}/ \Lambda$ are exactly the points, \(z = \frac{a}{n} \omega_1 + \frac{b}{n} \omega_2 \text{ for } a,b \in \mathbb{Z}/ n \mathbb{Z}\) such that $n z = a \omega_1 + b \omega_2 \in \Lambda$ is trivial in $\mathbb{C}/ \Lambda$. Therefore, $E[n] \cong (\mathbb{Z}/ n \mathbb{Z}) \oplus (\mathbb{Z}/ n \mathbb{Z})$ as abstract groups. ◻ :::

The Tate Module

::: proposition Proposition 19 . Let $E$ be an elliptic curve defined over $K$ then there is a natural action of the absolute Galois group, \(\mathrm{Gal} \small(\bar{K} / K \small) \to \mathrm{Aut} \small(E \small)\) which, because automorphism preserve $n$-torsion, reduces to an action, \(\mathrm{Gal} \small(\bar{K} / K \small) \to \mathrm{Aut} \small(E[n] \small) \cong \mathrm{GL}_{2}(\mathbb{Z}/ n \mathbb{Z})\) :::

::: proof Proof. Let $\sigma \in \mathrm{Gal} \small(\bar{K} / K \small)$ act component-wise on $E$ (and fix $O$). Because $\sigma$ is a field automorphism fixing $K$, the action of $\sigma$ preserves the defining equations of $E$ so it gives a map $E \to E$. Furthermore, since addition in $E$ is given by rational functions with coefficients in $K$, the action of $\sigma$ also preserves addition, that is, \(\sigma(P + Q) = \sigma(P) + \sigma(Q)\) and thus $\sigma$, being inevitable point-wise, is an automorphism of $E$. Finally, let $P_1, P_2$ be a $\mathbb{Z}/ n \mathbb{Z}$-basis of $E[n]$ as a free $\mathbb{Z}/ n \mathbb{Z}$-module. Then there exist unique elements $a, b, c, d \in \mathbb{Z}/ n \mathbb{Z}$ such that, \(\sigma(P_1) = a P_1 + c P_2 \text{ and } \sigma(P_2) = b P_1 + d P_2\) Then the action of $\sigma$ on any element $n_1 P_1 + n_2 P_2$ is given by, \(\sigma(n_1 P_1 + n_2 P_2) = n_1 \sigma(P_1) + n_2 \sigma(P_2) = (a n_1 + b n_2) P_1 + (c n_1 + d n_2) P_2 = n_1' P_1 + n_2' P_2\) Which we may write suggestively as, \(\begin{pmatrix} n_1' \\ n_2' \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} n_1 \\ n_2 \end{pmatrix}\) explicitly showing the representation of $\mathrm{Gal} \small(\bar{K} / K \small)$ as matrices in $\mathrm{GL}_{2}(\mathbb{Z}/ n \mathbb{Z})$. ◻ :::

::: trivlist The Tate module of an elliptic curve $E$ is the group, \(T_{\ell}(E) = \varprojlim E[\ell^n]\) under the multiplication by $\ell$ maps,

The Tate module $T_{\ell}(E)$ can be given the structure of a $\mathbb{Z} {\ell}$ module via the action, $(a_n) \in \mathbb{Z} {\ell}$ acts on $(P_n) \in T_{\ell}(E)$ via $(a_n) \cdot (P_n) = (a_n \cdot P_n)$. This action is well defined because $P_n$ has $\ell^n$ torsion so $a_n$ need only be defined up to multiples of $\ell^n$. :::

::: {#thm:galois_rep_tate_module .theorem} Theorem 20 . Let $E$ be an elliptic curve over $K / \mathbb{Q}$. There exists an $\ell$-adic Galois representation $V_{\ell} E = T_{\ell}(E) \otimes_{\mathbb{Z} {\ell}} \mathbb{Q} {\ell}$ with action, \(\rho_{E,\ell} : \mathrm{Gal} \small(\bar{K}/K \small) \to \mathrm{Aut} \small(V_{\ell}E \small) \cong \mathrm{GL}_{2}(\mathbb{Q}_{\ell})\) called the Galois representation attached to $E$ at $\ell$. :::

::: proof Proof. For each $n \in \mathbb{Z}^{+}$ we have an action of $\sigma \in \mathrm{Gal} \small(\bar{K} / K \small)$ on $P \in E[\ell^n]$ component-wise. However, $\ell \cdot \sigma(P) = \sigma(\ell \cdot P)$ because $\sigma$ is a group homomorphism of $E$ so $\sigma$ is compatible with the restriction maps. Therefore, $\sigma$ lifts to $\tilde{\sigma}$ a unique automorphism of the Tate module $T_{\ell}(E)$. By choosing bases compatible with the multiplication by $\ell$ maps gives an isomorphism, \(T_{\ell}(E) \cong \varprojlim{(\mathbb{Z}/ \ell^n \mathbb{Z}) \oplus (\mathbb{Z}/ \ell^n \mathbb{Z})} \cong \mathbb{Z}_{\ell} \oplus \mathbb{Z}_{\ell}\) The action on the Tate module induces a map, \(\rho_T : \mathrm{Gal} \small(\bar{K}/K \small) \to \mathrm{Aut} \small(T_{\ell}(E) \small) \cong \mathrm{Aut} \small(\mathbb{Z}_{\ell} \oplus \mathbb{Z}_{\ell} \small) = \mathrm{GL}_{2}(\mathbb{Z}_{\ell}) \subset \mathrm{GL}_{2}(\mathbb{Q}_{\ell})\) The desired map is given by taking the tensor product with the trivial $\ell$-adic representation $(\mathbb{Q} {\ell}, \rho_0)$, \(\rho_{E, \ell} = \rho_T \otimes \rho_0 : \mathrm{Gal} \small(\bar{K}/K \small) \to \mathrm{Aut} \small(T_{\ell} \otimes_{\mathbb{Z}_{\ell}} \mathbb{Q}_{\ell} \small) \cong \mathrm{Aut} \small(\mathbb{Q}_{\ell} \otimes \mathbb{Q}_{\ell} \small) = \mathrm{GL}_{2}(\mathbb{Q}_{\ell})\) and we take the $\mathbb{Q} {\ell}$ vector space, \(V_{\ell} E = T_{\ell} \otimes_{\mathbb{Z}_{\ell}} \mathbb{Q}_{\ell} \cong (\mathbb{Z}_{\ell} \oplus \mathbb{Z}_{\ell}) \otimes_{\mathbb{Z}_{\ell}} \mathbb{Q}_{\ell} = \mathbb{Q}_{\ell} \oplus \mathbb{Q}_{\ell}\) ◻ :::

::: remark Remark 8 . We have seen here an interesting $\ell$-adic representation arising naturally from algebraic geometry. It is a truly remarkable fact that a very large class of all $\ell$-adic Galois representations arise from geometric objects. :::

Complex Multiplication

We have discussed how $\ell$-adic representations can give information about the underlying geometry. Galois representations can also be used to determine algebraic properties of Galois groups from geometric structures.

::: trivlist An elliptic curve $E$ has complex multiplication, is CM for short, if there exists an endomorphism of $E$ which is not a multiplication by $n$ map. :::

::: remark Remark 9 . From the complex analytic viewpoint, such a map is multiplication by $c \in \mathbb{C}$ hence the name. Note, an endomorphism of an elliptic curve is required to be an isogeny i.e. given by rational functions over $\bar{K}$. :::

::: lemma Lemma 21 . Let $E$ be an elliptic defined over $K$ with complex multiplication and denote the exceptional endomorphism by $\phi : E \to E$. There exists a finite extension $K^{\mathrm{CM}} / K$ such that $\forall \sigma \in \mathrm{Gal} \small(\bar{K} / K^{\mathrm{CM}} \small) : \phi \circ \sigma = \sigma \circ \phi$ :::

::: proof Proof. Because $\phi$ is an isogeny, it is given by rational functions with coefficients in $\bar{K}$. Let $K^{\mathrm{CM}} = K(S)$ where $S$ is the set of coefficients of $\phi$. Then for any automorphism $\sigma \in \mathrm{Gal} \small(\bar{K} / K^{\mathrm{CM}} \small)$ we know that $\sigma$ preserves the coefficients of $\phi$ and thus, because $\sigma$ is a field homomorphism and $\phi$ is a rational function, $\sigma \circ \phi(P) = \phi \circ \sigma(P)$ for any point $P \in E$. ◻ :::

::: trivlist The field $K^{\mathrm{CM}}(E[n])$ is the field extension of $K^{\mathrm{CM}}$ generated by the coordinates of all points in $E[n]$. Furthermore define the $\ell^\infty$-torsion, \(E[\ell^\infty] = \bigcup_{n} E[\ell^n]\) and the field, \(K^{\mathrm{CM}}(E[\ell^\infty]) = \bigcup_{n} K^{\mathrm{CM}} (E[\ell^n]) = \varinjlim K^{\mathrm{CM}}(E[\ell^n])\) :::

::: lemma Lemma 22 . For each $n \in \mathbb{Z}^{+}$ the extension $K^{\mathrm{CM}} (E[\ell^n]) / K^{\mathrm{CM}}$ is finite Galois and therefore $K^{\mathrm{CM}} (E[\ell^\infty]) / K^{\mathrm{CM}}$ is Galois. :::

::: proof Proof. Take $\sigma \in \mathrm{Gal} \small(\bar{K} / K^{\mathrm{CM}} \small)$ and consider $\sigma(K^{\mathrm{CM}} (E[\ell^n]))$. Since $\sigma$ fixes $K^{\mathrm{CM}}$ the image of $K^{\mathrm{CM}} (E[\ell^n])$ is entirely determined by where $\sigma$ maps the generators which are coordinates of $E[\ell^n]$. However, we have shown that $\sigma$ is an automorphism of $E$ and thus must take $\ell^n$-torsion to $\ell^n$-torsion. Therefore, for each $P \in E[\ell^n]$ we have $\sigma(P) \in E[\ell^n]$ so $\sigma$ must map coordinates of $E[\ell^n]$ to coordinates of $E[\ell^n]$. Therefore, $\sigma(K^{\mathrm{CM}} (E[\ell^n])) = K^{\mathrm{CM}} (E[\ell^n])$ proving that $K^{\mathrm{CM}} (E[\ell^n])$ is Galois over $K^{\mathrm{CM}}$. Furthermore, since $E[\ell^n]$ is finite, there are only finitely many possible permutations and thus finitely map automorphisms of $K^{\mathrm{CM}} (E[\ell^n])$ proving the extension is finite. ◻ :::

::: theorem Theorem 23 . Let $E$ be an elliptic curve defined over $K$ with complex multiplication $\phi : E \to E$ such that $\phi$ restricted to $E[\ell]$ is not the multiplication by $n$ map for any $n \in \mathbb{Z}$. Then the extension $K^{\mathrm{CM}}(E[\ell^\infty]) / K^{\mathrm{CM}}$ is abelian. :::

::: proof Proof. Restricting the map given in Theorem 20 {reference-type=”ref” reference=”thm:galois_rep_tate_module”} gives a representation, \(\rho^{\mathrm{CM}}_{E, \ell} : \mathrm{Gal} \small(\bar{K} / K^{\mathrm{CM}} \small) \to \mathrm{Aut} \small(V_{\ell}E \small) \cong \mathrm{GL}_{2}(\mathbb{Q}_{\ell})\) We have that $\rho^{\mathrm{CM}} {E, \ell}(\sigma) = I \iff \sigma \cdot (P_n) = (\sigma(P_n)) = (P_n)$ for every $(P_n) \in T {\ell}(E)$. Therefore, $\sigma \in \ker{\rho^{\mathrm{CM}}_{E, \ell}}$ if and only if $\sigma$ acts trivially on $E[\ell^n]$ for each $n \in \mathbb{Z}^{+}$ or equivalently, since $\sigma$ acts coordinate-wise on $E[\ell^n]$, acting trivially on the coordinates of $E[\ell^n]$. However, acting trivially on the generators is equivalent to fixing the field $K^{\mathrm{CM}}(E[\ell^n])$ for all $n$ or, equivalently, fixing the compositum $K^{\mathrm{CM}}(E[\ell^\infty])$. Therefore, \(\ker{\rho^{\mathrm{CM}}_{E,\ell}} = \mathrm{Gal} \small(\bar{K} / K^{\mathrm{CM}}(E[\ell^\infty]) \small)\) Furthermore, the kernel is closed (because the action is continuous) and normal so the quotient, \(\mathrm{Gal} \small(\bar{K} / K^{\mathrm{CM}} \small) / \mathrm{Gal} \small(\bar{K} / K^{\mathrm{CM}}(E[\ell^\infty]) \small) \cong \mathrm{Gal} \small(K^{\mathrm{CM}}(E[\ell^\infty]) / K^{\mathrm{CM}} \small)\) corresponds to the Galois extension $K^{\mathrm{CM}}(E[\ell^\infty]) / K^{\mathrm{CM}}$. The action then injectivly factors through the quotient as,

so the group $\mathrm{Gal} \small(K^{\mathrm{CM}}(E[\ell^\infty]) / K^{\mathrm{CM}} \small)$ is embedded in $\mathrm{GL} {2}(\mathbb{Q} {\ell})$. However, $\sigma \circ \phi = \phi \circ \sigma$ for all $\sigma \in \mathrm{Gal} \small(\bar{K} / K \small)$ and $\phi$ is not multiplication by $n$ on $E[\ell]$ so $\phi \in \mathrm{Aut} \small(V_{\ell}E \small) \cong \mathrm{GL} {2}(\mathbb{Q} {\ell})$ corresponds to a non-scalar matrix. However, all matrices in $\mathrm{GL} {2}(\mathbb{Q} {\ell})$ which commute with a fixed non-scalar matrix (which remains non-scalar in the reduction modulo $\ell$) commute with each other (technical exercise). Therefore, the image of $\mathrm{Gal} \small(K^{\mathrm{CM}}(E[\ell^\infty]) / K^{\mathrm{CM}} \small)$ in $\mathrm{GL} {2}(\mathbb{Q} {\ell})$ is abelian so by the embedding $\mathrm{Gal} \small(K^{\mathrm{CM}}(E[\ell^\infty]) / K^{\mathrm{CM}} \small)$ is abelian itself. ◻ :::

::: example Example 3 . Consider the elliptic curve over $\mathbb{Q}$ defined by, \(E : y^2 = x^3 + x\) which has an exceptional automorphism $\phi : E \to E$ given by, \(\phi(x, y) = (-x, iy)\) which preserves the defining equation and the group law. Clearly, $K^{\mathrm{CM}} = \mathbb{Q}(i)$ since $i$ is the only non-rational coefficient defining $\phi$. One can easily check that $\phi$ is not multiplication by $n$ on any torsion subgroup. Therefore, the extensions \(\mathbb{Q}(i)(E[\ell^\infty]) / \mathbb{Q}(i)\) given by adjoining $\ell^n$-torsion points of $E$ are abelian for each $\ell$. :::

::: remark Remark 10 . This is an example of Kronecker’s Jugendtraum or “Dream of Youth” which was to generate all abelian extensions of a number field $K$ by adjoining special values of certain interesting functions. For example, the Kronecker-Weber theorem does this for $K = \mathbb{Q}$ saying that the abelian extensions of $\mathbb{Q}$ are exactly the subfields of $\mathbb{Q}(f(\tfrac{1}{n}))$ where $f(x) = e^{2 \pi i x}$ is a very special analytic function. It turns out that, astonishingly, our above construction generated all the abelian extensions of $\mathbb{Q}(i)$. That is, the compositum over all primes $\ell$ of $\mathbb{Q}(i)(E[\ell^\infty])$ gives the maximal abelian extension of $\mathbb{Q}(i)$. Equivalently, every finite abelian extension of $\mathbb{Q}(i)$ is contained in $\mathbb{Q}(i)$ adjoined some finite set of torsion points on the curve $E$. The theory of elliptic curves using Galois representations has now realized Kronecker’s dream for all imaginary quadratic fields. However, the general case remains a mystery. :::

Further Reading

Gradient Descent Done Right

Upmath: Math Online Editor Create web articles and blog posts with equations and diagrams Upmath extremely simplifies this task by using Markdown and LaTeX. It converts the Markdown syntax extende...

Berry Phase and Chern Classes

HI $c_1$ vs \[c_1\]

Walipiri Kinship Relations, Group Theory, and Reflections on Ethnomathematics

DO THIS Now this entire discussion has treated the Kinship problem as basically a curiosity. A curiosity to be understood and examined in the context of a great human persuit but still a curiosity...

A new version of content is available.

Stack Exchange Network

Stack Exchange network consists of 183 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

Understanding absolute Galois group from its representations

Background. A major theme of modern number theory is to study the absolute Galois group $\text{Gal}(\overline{\mathbb Q} / \mathbb Q)$ . Galois representation theory attempts to understand $\text{Gal}(\overline{\mathbb Q} / \mathbb Q)$ via its representations. Here is one example that I just learned. Given an elliptic curve $E$ over $\mathbb{Q}$ and a prime number $\ell$ , then there is a Galois representation given by: $$ \rho_{E, \ell} : \text{Gal}(\overline{\mathbb Q} / \mathbb Q) \rightarrow \text{Aut}(E[\ell]) \cong \text{GL}_2(\mathbb Z/\ell) $$ where $E[\ell]$ is the group of $\ell$ -torsion points of $E(\overline{\mathbb Q})$ . Serre conjectured that there is a constant $\ell_0$ such that for every $\ell > \ell_0$ and non-CM elliptic curve $E$ , the map $\rho_{E, \ell}$ is surjective. Now, all this makes me wonder as a "tourist" to number theory:

Question 1. What have mathematicians learned about $\text{Gal}(\overline{\mathbb Q} / \mathbb Q)$ itself through Galois representations? In other words, what are some major theorems from Galois representation theory and how did that help understand $\text{Gal}(\overline{\mathbb Q} / \mathbb Q)$ ?

Question 2. What are some major conjectures in Galois representation, and what does that make us believe about $\text{Gal}(\overline{\mathbb Q} / \mathbb Q)$ ?

I understand that there is a $p$ -adic version of Galois representation, which I believe is a local program to apply the local-global principle to understand $\text{Gal}(\overline{\mathbb Q} / \mathbb Q)$ . I'm happy to receive answers about $p$ -adic Galois representation as well.

(I am an applied mathematician with a keen interest in number theory)

• nt.number-theory
• galois-theory
• galois-representations

• 1 $\begingroup$ You wrote $\text{GL}_2(\mathbb{Z}/p)$ in your first display. Did you mean mod $\ell$? The letter $p$ is not mentioned prior to this display. This makes a big difference. $\endgroup$ –  Stanley Yao Xiao Commented Feb 12 at 21:40
• $\begingroup$ I edited that back to Z/ell, thank you! $\endgroup$ –  Uzu Lim Commented Feb 12 at 21:46

2 Answers 2

The slogan "number theorists aim to understand $\operatorname{Gal}(\overline{\mathbb{Q}} / \mathbb{Q})$ " is one that gets used a lot, but it's perhaps a tiny little bit misleading.

Understanding the structure of $\operatorname{Gal}(\overline{\mathbb{Q}} / \mathbb{Q})$ as an abstract group , or maybe as a topological group, is of course a highly interesting problem; but the general verdict is that it's a huge chaotic jungle. E.g. the Inverse Galois problem is the conjecture that every finite group occurs as the quotient of $\operatorname{Gal}(\overline{\mathbb{Q}} / \mathbb{Q})$ by an open subgroup, generally in infinitely many ways.

Studying "restricted ramification" Galois groups -- quotients of $\operatorname{Gal}(\overline{\mathbb{Q}} / \mathbb{Q})$ parametrizing extensions in which only a given finite set of primes is allowed to ramify -- cuts the jungle down to a managable size. These groups have tolerably explicit presentations in terms of (topological) generators and relations, as do the local groups $\operatorname{Gal}(\overline{\mathbb{Q}}_p / \mathbb{Q}_p)$ . But these abstract algebraic questions are somewhat disjoint from the mainstream of Galois-representation theory.

A more accurate, although much wordier, description of the main preoccupations of Galois representation theory might go as follows.

Galois representations "arise naturally" from arithmetically interesting objects (elliptic curves, modular forms, algebraic varieties).

Studying the Galois representations associated to these objects often provides a very powerful way of solving problems about their arithmetic, although those questions may initially seem to have nothing to do with Galois representations. Kolyvagin's proof of the Birch--Swinnerton-Dyer conjecture in analytic rank $\le 1$ is a classic example of this.

Studying the Galois representations gives a uniform way of understanding the arithmetic of many different classes of objects at the same time.

Conversely, all Galois representations which are "nice enough" -- expressed in terms of local conditions -- seem to arise from algebraic varieties (the Fontaine--Mazur conjecture).

• $\begingroup$ Your "wordier" summary is very helpful :) Thank you. I wrote a comment here before thanking you, but I think someone deleted it because I was simply "thanking" and not adding to the discussion, or whatever. Which is annoying, but fine, heh $\endgroup$ –  Uzu Lim Commented Feb 13 at 22:45

$\newcommand{\Q}{\mathbb{Q}} \DeclareMathOperator{\Gal}{Gal}$ Here is an example of theorem that may be in the style that you are looking for, in the sense that the statement does not involve Galois representations but its proof does. On the other hand, I think it illustrates well David Loeffler's point that number theorists are often interested in the Galois group not as an abstract topological group, but together with the ramification information coming from the completions of $\Q$ .

Theorem (Chenevier - Clozel): Let $S$ be a finite set of primes, let $p\in S$ and let $\Q_S$ be the maximal algebraic extension of $\Q$ unramified outside of $S$ . Assume $|S|\ge 2$ . Then the natural map $$ \Gal(\bar{\Q}_p/\Q_p) \to \Gal(\Q_S/\Q) $$ is injective.

Note that the corresponding statement with $\Q_S$ replaced by $\bar\Q$ is easy and does not require any Galois representations at all!

Reference: G. Chenevier and L. Clozel, Corps de nombres peu ramifiés et formes automorphes autoduales , J. Amer. Math. Soc., 2009.

• $\begingroup$ So perhaps when people say "we want to understand the absolute Gal group", they also mean to understand the action on other arithmetic objects. Which happen to be vector spaces anyway. Thank you! $\endgroup$ –  Uzu Lim Commented Feb 13 at 16:44
• 1 $\begingroup$ This is an excellent example. However, as far as I'm aware, it's also quite an unusual one: I'd go as far as to say that it's the only example I know of where automorphic forms and other heavyweight machinery from the Langlands program has been used to prove a result about Galois groups which is (almost) purely group-theoretic. $\endgroup$ –  David Loeffler Commented Feb 13 at 21:38
• $\begingroup$ @UzuLim Was that meant to be a comment on Aurel's answer, or on mine? $\endgroup$ –  David Loeffler Commented Feb 13 at 21:38
• $\begingroup$ @DavidLoeffler I agree it is unusual. There are also a few cases of the inverse Galois problem too, but the relation to Galois representations is more obvious. And Gross's problem on the unsolvability of $Gal(\mathbb{Q}_S/\mathbb{Q})$. But that's about all the examples I know. $\endgroup$ –  Aurel Commented Feb 13 at 22:02

Your Answer

Sign up or log in, post as a guest.

Required, but never shown

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy .

Not the answer you're looking for? Browse other questions tagged nt.number-theory galois-theory galois-representations or ask your own question .

• Featured on Meta
• Upcoming initiatives on Stack Overflow and across the Stack Exchange network...
• Announcing a change to the data-dump process

Get the Reddit app

A subreddit for those who enjoy learning about flags, their place in society past and present, and their design characteristics

The flag of Elektrostal, Moscow Oblast, Russia which I bought there during my last visit

By continuing, you agree to our User Agreement and acknowledge that you understand the Privacy Policy .

Enter the 6-digit code from your authenticator app

You’ve set up two-factor authentication for this account.

Enter a 6-digit backup code

Create your username and password.

Reddit is anonymous, so your username is what you’ll go by here. Choose wisely—because once you get a name, you can’t change it.

Reset your password

Enter your email address or username and we’ll send you a link to reset your password

Check your inbox

An email with a link to reset your password was sent to the email address associated with your account

Choose a Reddit account to continue

Stack Exchange Network

Stack Exchange network consists of 183 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

Q&A for work

Connect and share knowledge within a single location that is structured and easy to search.

On a certain representation of the Galois group of $X^n-a$ from Lang's Algebra [duplicate]

I am having trouble understand theorem $9.4$ of Chapter $6$ of Lang's Algebra (pg. 300-301).

The setup is a we have a field $k$ of characteristic not dividing $n$. We know that the splitting field of $f=X^n-a$ is $k(\zeta_n,\alpha)$ where $\alpha$ is a root of $f$ and $\zeta$ a primitive $n^{th}$ root of unity. Any automorphism $\sigma$ of the Galois group of $f$ over $k$ maps $\alpha \mapsto \alpha\zeta^b$ where $b$ is unique modulo $n$, and $\sigma$ induces an automorphism of the cyclic group $\mathbf{\mu_n}=\left\langle \zeta \right \rangle $ via $\zeta \mapsto \zeta^d$ where $(d,n)=1$ and $d$ is uniquely determined by $\sigma$.

We then verify that the map $\sigma \mapsto \begin{pmatrix} \ \ 1 & 0\\ b_\sigma & d_\sigma \end{pmatrix}$ where $b$, $d$ are the integers determined by $\sigma$ in the previous paragraph is an injective homomorphism into the group $G(n)$ of all matrices $\begin{pmatrix} \ 1 & 0\\ a & c \end{pmatrix}$ such that $a \in \mathbf{Z}/n\mathbf{Z}$, $c \in (\mathbf{Z}/n\mathbf{Z})^{*}.$

The question that the theorem addresses is when the above the map is an isomorphism of the Galois group of $f$ and $G(n)$. With $\phi$ being the Euler function, the theorem states:

Suppose $[k(\zeta_n):k]=\phi(n)$ and let $a \in k$. Suppose for each prime $p|n$ that $a$ is not a $p^{th}$ power. Let $K$ be the splitting field of $X^n-a$ over $k$ and $G$ the Galois group. Then the above map is an isomorphism $G \cong G(n)$ with commutator subgroup Gal$(K/k(\zeta_n))$, so $k(\zeta_n)$ is the maximal abelian subextension of $K$.

The proof begins with the case $n=p$ where $p$ is a prime, which I follow. However, following that case Lang writes (bold is what I don't understand):

A direct computation of commutator of elements in $G(n)$ for arbitrary n shows that the commutator subgroup $C$ is contained in the group of matrices $ \begin{pmatrix} \ \ 1 & 0\\ b & 1 \end{pmatrix}$, $b \in \mathbf{Z}/n\mathbf{Z}$ and so must be that subgroup because its factor group is isomorphic to $(\mathbf{Z}/n\mathbf{Z})^{*}$ under the projection on the diagonal .

When $n=p$ is prime I already happened to know that $G \cong \mathbf{Z}/p\mathbf{Z} \rtimes_\varphi (\mathbf{Z}/p\mathbf{Z})^{*}$ from which it is clear (I think?) that the quotient by the image of $\mathbf{Z}/p\mathbf{Z}$ is the maximal abelian quotient, and the fact that the commutator subgroup is nontrivial inside a subgroup of order $p$ means it must be the whole group. However, when $n$ is arbitrary it's not obvious to me why the quotient by the commutator $C$ in $G(n)$ is ismorphic to $(\mathbf{Z}/n\mathbf{Z})^{*}$ nor why $C$ has to be isomorphic to $\mathbf{Z}/n\mathbf{Z}$. If someone could explain what is missing here that would be much appreciated.

I also have a couple of questions on the rest of the argument which I will just link here

rest of proof

1.) On the $3^{rd}-4^{th}$ lines: $\beta$ is a root of $X^m-a$ and by induction we can apply the theorem to $g=X^m-a$ .

OK, fine, but what is Lang using the induction for? The splitting field for $g$ is $k(\beta,\zeta_m)$ and has Galois group isomorphic to $G(m)$ and it's maximal abelian extension is $k(\zeta_m)$. I've been staring at this and I don't see what the conclusion is.

2.) lines $3-5$ after the diagram: apply the $1^{st}$ part of proof (case of $n=p$ is prime) to $X^p-\beta$ over $k(\beta)$...shows that $k(\beta,\zeta_n)\cap k(\alpha)=k(\beta)$ .

Again I don't know what exactly is being said. The splitting field of $X^p-\beta$ over $k(\beta)$ is $k(\alpha,\zeta_p)$, and it's maximal abelian subextension is $k(\beta,\zeta_p)$. How does Lang's conclusion follow from that?

Thanks to everyone who took the time to read my question, any help is much appreciated.

• abstract-algebra
• field-theory
• galois-theory
• galois-representations
• kummer-theory

• $\begingroup$ As far as I know, it's so-called generalized Kummer theory, which is related to Galois representations (I haven't dug into this area, sorry). I don't know whether this material is clearer. $\endgroup$ –  Yai0Phah Commented Nov 25, 2015 at 20:09

I have given a complete proof of the Theorem in my own post here . The OP will probably not profit from it but I hope some desperate algebra students will.

Not the answer you're looking for? Browse other questions tagged abstract-algebra field-theory galois-theory galois-representations kummer-theory .

• Featured on Meta
• Upcoming initiatives on Stack Overflow and across the Stack Exchange network...
• Announcing a change to the data-dump process

Hot Network Questions

• How can dragons breathe in space?
• Would it be possible to generate data from real data in medical research?
• What's the mathematical idea behind weather forecasting POP (probability of precipitation) by slices of time?
• Is every Boolean algebra a subalgebra of a free one?
• Is there a way to render someone braindead while leaving the organs and body fully intact?
• Reduce a string up to idempotency
• Why "praemiīs mē dōnant" and not "praemia mihi donant"?
• How is Agile model more flexible than the Waterfall model?
• Integer solutions for a simple cubic
• Has any US president ever endorsed Trump for US presidency?
• In this position, why is the move 12. Be3 rarely played by top players?
• How to create arrowheads in this curve in Mathematica
• Civic Responsibility on London Railway regarding others tailgating us
• Travelling from Ireland to Northern Ireland (UK Visa required national)
• Proper explanation of why do we need learning rate in gradient descent?
• Take screenshot of rendered HTML from Linux without GUI
• With SHA-1 broken, can the old Linux /dev/(u)random algorithm be trusted?
• Why does MoleculeModify only accept a single modification?
• Number of leaves in complete binary tree
• Simple instance illustrating significance of Langlands dual group without getting into the Langlands program?
• "Nam sī qua alia urbs, est īnsalūbris Rōma."
• Can true ever be false?
• What's the frequency level of 時々?
• A hat puzzle question—how to prove the standard solution is optimal?